Question

The total number of natural numbers of six digits that can be made with digits $1,2,3,4$, if all digits are to appear in the same number at least once, is

• A. 1560
• B. 840
• C. 1080
• D. 480

Answer 1

Answer: (A)

There can be two types of numbers:
(i) Any one of the digits $1,2,3,4$ appears thrice and the remaining digits only once, that is, of the type 1,2 , $3,4,4,4$, etc. Number of ways of selection of digit which appears thrice $={}^4{C}_1$
Then, number of numbers of this type
$=\frac{6!}{3!}\times {}^4{C}_1=480$
(ii) Any two of the digits $1,2,3,4$ appears twice and the remaining two only once, that is, of the type $1,2,3$, $3,4,4$, etc.
Number of ways of selection of two digits which appears twice $={}^4{C}_2$
Then, number of numbers of this type
$=\frac{6!}{2!2!}\times {}^4{C}_2=1080$
Therefore, the required number of numbers $=480+$ $1080=1560$